Engineering Mathematics 9/2012 ppt

Engineering Mathematics
In memory of Elizabeth
Engineering Mathematics
Fourth Edition
JOHN BIRD, BSc(Hons) CMath, FIMA, CEng, MIEE, FCollP, FIIE
Newnes
OXFORD AMSTERDAM BOSTON LONDON NEW YORK PARIS
SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO
Newnes
An imprint of Elsevier Science
Linacre House, Jordan Hill, Oxford OX2 8DP
200 Wheeler Road, Burlington MA 01803
First published 1989
Second edition 1996
Reprinted 1998 (twice), 1999
Third edition 2001
Fourth edition 2003
Copyright
 2001, 2003, John Bird. All rights reserved
The right of John Bird to be identified as the author of this work
has been asserted in accordance with the Copyright, Designs and
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Contents
Preface xi
Part 1 Number and Algebra 1
1 Revision of fractions, decimals and
percentages 1
1.1 Fractions 1
1.2 Ratio and proportion 3
1.3 Decimals 4
1.4 Percentages 7
2 Indices and standard form 9
2.1 Indices 9
2.2 Worked problems on indices 9
2.3 Further worked problems on
indices 11
2.4 Standard form 13
2.5 Worked problems on standard
form 13
2.6 Further worked problems on standard

form 14
3 Computer numbering systems 16
3.1 Binary numbers 16
3.2 Conversion of binary to decimal 16
3.3 Conversion of decimal to binary 17
3.4 Conversion of decimal to binary via
octal 18
3.5 Hexadecimal numbers 20
4 Calculations and evaluation of
formulae 24
4.1 Errors and approximations 24
4.2 Use of calculator 26
4.3 Conversion tables and charts 28
4.4 Evaluation of formulae 30
Assignment 1 33
5 Algebra 34
5.1 Basic operations 34
5.2 Laws of Indices 36
5.3 Brackets and factorisation 38
5.4 Fundamental laws and precedence 40
5.5 Direct and inverse proportionality 42
6 Further algebra 44
6.1 Polynomial division 44
6.2 The factor theorem 46
6.3 The remainder theorem 48
7 Partial fractions 51
7.1 Introduction to partial fractions 51
7.2 Worked problems on partial fractions
with linear factors 51
7.3 Worked problems on partial fractions

with repeated linear factors 54
7.4 Worked problems on partial fractions
with quadratic factors 55
8 Simple equations 57
8.1 Expressions, equations and
identities 57
8.2 Worked problems on simple
equations 57
8.3 Further worked problems on simple
equations 59
8.4 Practical problems involving simple
equations 61
8.5 Further practical problems involving
simple equations 62
Assignment 2 64
9 Simultaneous equations 65
9.1 Introduction to simultaneous
equations 65
9.2 Worked problems on simultaneous
equations in two unknowns 65
9.3 Further worked problems on
simultaneous equations 67
9.4 More difficult worked problems on
simultaneous equations 69
9.5 Practical problems involving
simultaneous equations 70
10 Transposition of formulae 74
10.1 Introduction to transposition of
formulae 74
10.2 Worked problems on transposition of

formulae 74
10.3 Further worked problems on
transposition of formulae 75
10.4 Harder worked problems on
transposition of formulae 77
11 Quadratic equations 80
11.1 Introduction to quadratic equations 80
11.2 Solution of quadratic equations by
factorisation 80
vi CONTENTS
11.3 Solution of quadratic equations by
‘completing the square’ 82
11.4 Solution of quadratic equations by
formula 84
11.5 Practical problems involving quadratic
equations 85
11.6 The solution of linear and quadratic
equations simultaneously 87
12 Logarithms 89
12.1 Introduction to logarithms 89
12.2 Laws of logarithms 89
12.3 Indicial equations 92
12.4 Graphs of logarithmic functions 93
Assignment 3 94
13 Exponential functions 95
13.1 The exponential function 95
13.2 Evaluating exponential functions 95
13.3 The power series for e
x
96

13.4 Graphs of exponential functions 98
13.5 Napierian logarithms 100
13.6 Evaluating Napierian logarithms 100
13.7 Laws of growth and decay 102
14 Number sequences 106
14.1 Arithmetic progressions 106
14.2 Worked problems on arithmetic
progression 106
14.3 Further worked problems on arithmetic
progressions 107
14.4 Geometric progressions 109
14.5 Worked problems on geometric
progressions 110
14.6 Further worked problems on geometric
progressions 111
14.7 Combinations and permutations 112
15 The binomial series 114
15.1 Pascal’s triangle 114
15.2 The binomial series 115
15.3 Worked problems on the binomial
series 115
15.4 Further worked problems on the
binomial series 117
15.5 Practical problems involving the
binomial theorem 120
16 Solving equations by iterative
methods 123
16.1 Introduction to iterative methods 123
16.2 The Newton–Raphson method 123
16.3 Worked problems on the

Newton–Raphson method 123
Assignment 4 126
Multiple choice questions on chapters 1 to
16 127
Part 2 Mensuration 131
17 Areas of plane figures 131
17.1 Mensuration 131
17.2 Properties of quadrilaterals 131
17.3 Worked problems on areas of plane
figures 132
17.4 Further worked problems on areas of
plane figures 135
17.5 Worked problems on areas of
composite figures 137
17.6 Areas of similar shapes 138
18 The circle and its properties 139
18.1 Introduction 139
18.2 Properties of circles 139
18.3 Arc length and area of a sector 140
18.4 Worked problems on arc length and
sector of a circle 141
18.5 The equation of a circle 143
19 Volumes and surface areas of
common solids 145
19.1 Volumes and surface areas of
regular solids 145
19.2 Worked problems on volumes and
surface areas of regular solids 145
19.3 Further worked problems on volumes
and surface areas of regular

solids 147
19.4 Volumes and surface areas of frusta of
pyramids and cones 151
19.5 The frustum and zone of a sphere 155
19.6 Prismoidal rule 157
19.7 Volumes of similar shapes 159
20 Irregular areas and volumes and mean
values of waveforms 161
20.1 Areas of irregular figures 161
20.2 Volumes of irregular solids 163
20.3 The mean or average value of a
waveform 164
Assignment 5 168
Part 3 Trigonometry 171
21 Introduction to trigonometry 171
21.1 Trigonometry 171
21.2 The theorem of Pythagoras 171
21.3 Trigonometric ratios of acute
angles 172
CONTENTS vii
21.4 Fractional and surd forms of
trigonometric ratios 174
21.5 Solution of right-angled triangles 175
21.6 Angles of elevation and
depression 176
21.7 Evaluating trigonometric ratios of any
angles 178
21.8 Trigonometric approximations for small
angles 181
22 Trigonometric waveforms 182

22.1 Graphs of trigonometric functions 182
22.2 Angles of any magnitude 182
22.3 The production of a sine and cosine
wave 185
22.4 Sine and cosine curves 185
22.5 Sinusoidal form A sinωt š ˛ 189
22.6 Waveform harmonics 192
23 Cartesian and polar co-ordinates 194
23.1 Introduction 194
23.2 Changing from Cartesian into polar
co-ordinates 194
23.3 Changing from polar into Cartesian
co-ordinates 196
23.4 Use of R ! P and P ! R functions on
calculators 197
Assignment 6 198
24 Triangles and some practical
applications 199
24.1 Sine and cosine rules 199
24.2 Area of any triangle 199
24.3 Worked problems on the solution of
triangles and their areas 199
24.4 Further worked problems on the
solution of triangles and their
areas 201
24.5 Practical situations involving
trigonometry 203
24.6 Further practical situations involving
trigonometry 205
25 Trigonometric identities and

equations 208
25.1 Trigonometric identities 208
25.2 Worked problems on trigonometric
identities 208
25.3 Trigonometric equations 209
25.4 Worked problems (i) on trigonometric
equations 210
25.5 Worked problems (ii) on trigonometric
equations 211
25.6 Worked problems (iii) on trigonometric
equations 212
25.7 Worked problems (iv) on trigonometric
equations 212
26 Compound angles 214
26.1 Compound angle formulae 214
26.2 Conversion of a sin ωt C b cos ωt into
R sinωt C ˛) 216
26.3 Double angles 220
26.4 Changing products of sines and cosines
into sums or differences 221
26.5 Changing sums or differences of sines
and cosines into products 222
Assignment 7 224
Multiple choice questions on chapters 17
to 26 225
Part 4 Graphs 231
27 Straight line graphs 231
27.1 Introduction to graphs 231
27.2 The straight line graph 231
27.3 Practical problems involving straight

line graphs 237
28 Reduction of non-linear laws to linear
form 243
28.1 Determination of law 243
28.2 Determination of law involving
logarithms 246
29 Graphs with logarithmic scales 251
29.1 Logarithmic scales 251
29.2 Graphs of the form y D ax
n
251
29.3 Graphs of the form y D ab
x
254
29.4 Graphs of the form y D ae
kx
255
30 Graphical solution of equations 258
30.1 Graphical solution of simultaneous
equations 258
30.2 Graphical solution of quadratic
equations 259
30.3 Graphical solution of linear and
quadratic equations simultaneously
263
30.4 Graphical solution of cubic equations
264
31 Functions and their curves 266
31.1 Standard curves 266
31.2 Simple transformations 268

31.3 Periodic functions 273
31.4 Continuous and discontinuous
functions 273
31.5 Even and odd functions 273
31.6 Inverse functions 275
Assignment 8 279
viii CONTENTS
Part 5 Vectors 281
32 Vectors 281
32.1 Introduction 281
32.2 Vector addition 281
32.3 Resolution of vectors 283
32.4 Vector subtraction 284
33 Combination of waveforms 287
33.1 Combination of two periodic
functions 287
33.2 Plotting periodic functions 287
33.3 Determining resultant phasors by
calculation 288
Part 6 Complex Numbers 291
34 Complex numbers 291
34.1 Cartesian complex numbers 291
34.2 The Argand diagram 292
34.3 Addition and subtraction of complex
numbers 292
34.4 Multiplication and division of complex
numbers 293
34.5 Complex equations 295
34.6 The polar form of a complex
number 296

34.7 Multiplication and division in polar
form 298
34.8 Applications of complex numbers 299
35 De Moivre’s theorem 303
35.1 Introduction 303
35.2 Powers of complex numbers 303
35.3 Roots of complex numbers 304
Assignment 9 306
Part 7 Statistics 307
36 Presentation of statistical data 307
36.1 Some statistical terminology 307
36.2 Presentation of ungrouped data 308
36.3 Presentation of grouped data 312
37 Measures of central tendency and
dispersion 319
37.1 Measures of central tendency 319
37.2 Mean, median and mode for discrete
data 319
37.3 Mean, median and mode for grouped
data 320
37.4 Standard deviation 322
37.5 Quartiles, deciles and percentiles 324
38 Probability 326
38.1 Introduction to probability 326
38.2 Laws of probability 326
38.3 Worked problems on probability 327
38.4 Further worked problems on
probability 329
38.5 Permutations and combinations 331
39 The binomial and Poisson distribution 333

39.1 The binomial distribution 333
39.2 The Poisson distribution 336
Assignment 10 339
40 The normal distribution 340
40.1 Introduction to the normal distribution
340
40.2 Testing for a normal distribution 344
41 Linear correlation 347
41.1 Introduction to linear correlation 347
41.2 The product-moment formula for
determining the linear correlation
coefficient 347
41.3 The significance of a coefficient of
correlation 348
41.4 Worked problems on linear
correlation 348
42 Linear regression 351
42.1 Introduction to linear regression 351
42.2 The least-squares regression lines 351
42.3 Worked problems on linear
regression 352
43 Sampling and estimation theories 356
43.1 Introduction 356
43.2 Sampling distributions 356
43.3 The sampling distribution of the
means 356
43.4 The estimation of population
parameters based on a large sample
size 359
43.5 Estimating the mean of a population

based on a small sample size 364
Assignment 11 368
Multiple choice questions on chapters 27
to 43 369
Part 8 Differential Calculus 375
44 Introduction to differentiation 375
44.1 Introduction to calculus 375
44.2 Functional notation 375
44.3 The gradient of a curve 376
44.4 Differentiation from first
principles 377
CONTENTS ix
44.5 Differentiation of y D ax
n
by the
general rule 379
44.6 Differentiation of sine and cosine
functions 380
44.7 Differentiation of e
ax
and ln ax 382
45 Methods of differentiation 384
45.1 Differentiation of common functions
384
45.2 Differentiation of a product 386
45.3 Differentiation of a quotient 387
45.4 Function of a function 389
45.5 Successive differentiation 390
46 Some applications of differentiation 392
46.1 Rates of change 392

46.2 Velocity and acceleration 393
46.3 Turning points 396
46.4 Practical problems involving maximum
and minimum values 399
46.5 Tangents and normals 403
46.6 Small changes 404
Assignment 12 406
Part 9 Integral Calculus 407
47 Standard integration 407
47.1 The process of integration 407
47.2 The general solution of integrals of the
form ax
n
407
47.3 Standard integrals 408
47.4 Definite integrals 411
48 Integration using algebraic substitutions
414
48.1 Introduction 414
48.2 Algebraic substitutions 414
48.3 Worked problems on integration using
algebraic substitutions 414
48.4 Further worked problems on integration
using algebraic substitutions 416
48.5 Change of limits 416
49 Integration using trigonometric
substitutions 418
49.1 Introduction 418
49.2 Worked problems on integration of
sin

2
x,cos
2
x,tan
2
x and cot
2
x 418
49.3 Worked problems on powers of sines
and cosines 420
49.4 Worked problems on integration of
products of sines and cosines 421
49.5 Worked problems on integration using
the sin  substitution 422
49.6 Worked problems on integration using
the tan  substitution 424
Assignment 13 425
50 Integration using partial fractions 426
50.1 Introduction 426
50.2 Worked problems on integration using
partial fractions with linear
factors 426
50.3 Worked problems on integration using
partial fractions with repeated linear
factors 427
50.4 Worked problems on integration using
partial fractions with quadratic
factors 428
51 The t
=

q
2
substitution 430
51.1 Introduction 430
51.2 Worked problems on the t D tan
Â
2
substitution 430
51.3 Further worked problems on the
t D tan
Â
2
substitution 432
52 Integration by parts 434
52.1 Introduction 434
52.2 Worked problems on integration by
parts 434
52.3 Further worked problems on integration
by parts 436
53 Numerical integration 439
53.1 Introduction 439
53.2 The trapezoidal rule 439
53.3 The mid-ordinate rule 441
53.4 Simpson’s rule 443
Assignment 14 447
54 Areas under and between curves 448
54.1 Area under a curve 448
54.2 Worked problems on the area under a
curve 449
54.3 Further worked problems on the area

under a curve 452
54.4 The area between curves 454
55 Mean and root mean square values 457
55.1 Mean or average values 457
55.2 Root mean square values 459
56 Volumes of solids of revolution 461
56.1 Introduction 461
56.2 Worked problems on volumes of solids
of revolution 461
x CONTENTS
56.3 Further worked problems on volumes
of solids of revolution 463
57 Centroids of simple shapes 466
57.1 Centroids 466
57.2 The first moment of area 466
57.3 Centroid of area between a curve and
the x-axis 466
57.4 Centroid of area between a curve and
the y-axis 467
57.5 Worked problems on centroids of
simple shapes 467
57.6 Further worked problems on centroids
of simple shapes 468
57.7 Theorem of Pappus 471
58 Second moments of area 475
58.1 Second moments of area and radius of
gyration 475
58.2 Second moment of area of regular
sections 475
58.3 Parallel axis theorem 475

58.4 Perpendicular axis theorem 476
58.5 Summary of derived results 476
58.6 Worked problems on second moments
of area of regular sections 476
58.7 Worked problems on second moments
of areas of composite areas 480
Assignment 15 482
Part 10 Further Number and Algebra 483
59 Boolean algebra and logic circuits 483
59.1 Boolean algebra and switching circuits
483
59.2 Simplifying Boolean expressions 488
59.3 Laws and rules of Boolean algebra
488
59.4 De Morgan’s laws 490
59.5 Karnaugh maps 491
59.6 Logic circuits 495
59.7 Universal logic circuits 500
60 The theory of matrices and determinants
504
60.1 Matrix notation 504
60.2 Addition, subtraction and multiplication
of matrices 504
60.3 The unit matrix 508
60.4 The determinant of a 2 by 2 matrix
508
60.5 The inverse or reciprocal of a 2 by 2
matrix 509
60.6 The determinant of a 3 by 3 matrix
510

60.7 The inverse or reciprocal of a 3 by 3
matrix 511
61 The solution of simultaneous equations by
matrices and determinants 514
61.1 Solution of simultaneous equations by
matrices 514
61.2 Solution of simultaneous equations by
determinants 516
61.3 Solution of simultaneous equations
using Cramers rule 520
Assignment 16 521
Multiple choice questions on chapters 44–61
522
Answers to multiple choice questions 526
Index 527
Preface
This fourth edition of ‘Engineering Mathematics’
covers a wide range of syllabus requirements.In
particular, the book is most suitable for the latest
National Certificate and Diploma courses and
Vocational Certificate of Education syllabuses in
Engineering.
This text will provide a foundation in mathematical
principles, which will enable students to solve mathe-
matical, scientific and associated engineering princi-
ples. In addition, the material will provide engineer-
ing applications and mathematical principles neces-
sary for advancement onto a range of Incorporated
Engineer degree profiles. It is widely recognised that
a students’ ability to use mathematics is a key element

in determining subsequent success. First year under-
graduates who need some remedial mathematics will
also find this book meets their needs.
In Engineering Mathematics 4
th
Edition, theory
is introduced in each chapter by a simple outline of
essential definitions, formulae, laws and procedures.
The theory is kept to a minimum, for problem solv-
ing is extensively used to establish and exemplify
the theory. It is intended that readers will gain real
understanding through seeing problems solved and
then through solving similar problems themselves.
For clarity, the text is divided into ten topic
areas, these being: number and algebra, mensura-
tion, trigonometry, graphs, vectors, complex num-
bers, statistics, differential calculus, integral calculus
and further number and algebra.
This new edition will cover the following syl-
labuses:
(i) Mathematics for Technicians, the core unit
for National Certificate/Diploma courses in
Engineering, to include all or part of the
following chapters:
1. Algebra: 2, 4, 5, 8–13, 17, 19, 27, 30
2. Trigonometry: 18, 21, 22, 24
3. Statistics: 36, 37
4. Calculus: 44, 46, 47, 54
(ii) Further Mathematics for Technicians,
the optional unit for National Certifi-

cate/Diploma courses in Engineering, to
include all or part of the following chapters:
1. Algebraic techniques: 10, 14, 15,
28–30, 34, 59–61
2. Trigonometry: 22–24, 26
3. Calculus: 44–49, 52–58
4. Statistical and probability:36–43
(iii) Applied Mathematics in Engineering,the
compulsory unit for Advanced VCE (for-
merly Advanced GNVQ), to include all or
part of the following chapters:
1. Number and units:1,2,4
2. Mensuration:17–20
3. Algebra: 5, 8–11
4. Functions and graphs: 22, 23, 27
5. Trigonometry: 21, 24
(iv) Further Mathematics for Engineering,the
optional unit for Advanced VCE (formerly
Advanced GNVQ), to include all or part of
the following chapters:
1. Algebra and trigonometry:5,6,
12–15, 21, 25
2. Graphical and numerical techniques:
20, 22, 26–31
3. Differential and integral calculus:
44–47, 54
(v) The Mathematics content of Applied Sci-
ence and Mathematics for Engineering,
for Intermediate GNVQ
(vi) Mathematics for Engineering,forFounda-

tion and Intermediate GNVQ
(vii) Mathematics 2 and Mathematics 3 for City
& Guilds Technician Diploma in Telecom-
munications and Electronic Engineering
(viii) Any introductory/access/foundation co-
urse involving Engineering Mathematics at
University, Colleges of Further and Higher
education and in schools.
Each topic considered in the text is presented in
a way that assumes in the reader little previous
knowledge of that topic.
xii ENGINEERING MATHEMATICS
‘Engineering Mathematics 4
th
Edition’ provides
a follow-up to ‘Basic Engineering Mathematics’
and a lead into ‘Higher Engineering Mathemat-
ics’.
This textbook contains over 900 worked
problems, followed by some 1700 further
problems (all with answers). The further problems
are contained within some 208 Exercises; each
Exercise follows on directly from the relevant
section of work, every two or three pages. In
addition, the text contains 234 multiple-choice
questions. Where at all possible, the problems
mirror practical situations found in engineering
and science. 500 line diagrams enhance the
understanding of the theory.
At regular intervals throughout the text are some

16 Assignments to check understanding. For exam-
ple, Assignment 1 covers material contained in
Chapters 1 to 4, Assignment 2 covers the material
in Chapters 5 to 8, and so on. These Assignments
do not have answers given since it is envisaged that
lecturers could set the Assignments for students to
attempt as part of their course structure. Lecturers’
may obtain a complimentary set of solutions of the
Assignments in an Instructor’s Manual available
from the publishers via the internet — full worked
solutions and mark scheme for all the Assignments
are contained in this Manual, which is available to
lecturers only. To obtain a password please e-mail
with the following details:
course title, number of students, your job title and
work postal address.
To download the Instructor’s Manual visit
and enter the book
title in the search box, or use the following direct
URL: />‘Learning by Example’ is at the heart of ‘Engi-
neering Mathematics 4
th
Edition’.
John Bird
University of Portsmouth
Part 1 Number and Algebra
1
Revision of fractions, decimals
and percentages
1.1 Fractions

When 2 is divided by 3, it may be written as
2
3
or
2/3.
2
3
is called a fraction. The number above the
line, i.e. 2, is called the numerator and the number
below the line, i.e. 3, is called the denominator.
When the value of the numerator is less than
the value of the denominator, the fraction is called
a proper fraction; thus
2
3
is a proper fraction.
When the value of the numerator is greater than
the denominator, the fraction is called an improper
fraction. Thus
7
3
is an improper fraction and can also
be expressed as a mixed number, that is, an integer
and a proper fraction. Thus the improper fraction
7
3
is equal to the mixed number 2
1
3
.

When a fraction is simplified by dividing the
numerator and denominator by the same number,
the process is called cancelling. Cancelling by 0 is
not permissible.
Problem 1. Simplify
1
3
C
2
7
The lowest common multiple (i.e. LCM) of the two
denominators is 3 ð7, i.e. 21
Expressing each fraction so that their denomina-
tors are 21, gives:
1
3
C
2
7
D
1
3
ð
7
7
C
2
7
ð
3

3
D
7
21
C
6
21
D
7 C 6
21
D
13
21
Alternatively:
1
3
C
2
7
D
Step (2)
#
7 ð1
C
Step (3)
#
3 ð2
21
"
Step (1)

Step 1: the LCM of the two denominators;
Step 2: for the fraction
1
3
, 3 into 21 goes 7 times,
7 ð the numerator is 7 ð1;
Step 3: for the fraction
2
7
, 7 into 21 goes 3 times,
3 ð the numerator is 3 ð2.
Thus
1
3
C
2
7
D
7 C 6
21
D
13
21
as obtained previously.
Problem 2. Find the value of 3
2
3
 2
1
6

One method is to split the mixed numbers into
integers and their fractional parts. Then
3
2
3
 2
1
6
D

3 C
2
3



2 C
1
6

D 3 C
2
3
 2 
1
6
D 1 C
4
6


1
6
D 1
3
6
D 1
1
2
Another method is to express the mixed numbers as
improper fractions.
2 ENGINEERING MATHEMATICS
Since 3 D
9
3
,then3
2
3
D
9
3
C
2
3
D
11
3
Similarly, 2
1
6
D

12
6
C
1
6
D
13
6
Thus 3
2
3
 2
1
6
D
11
3

13
6
D
22
6

13
6
D
9
6
D 1

1
2
as obtained previously.
Problem 3. Determine the value of
4
5
8
 3
1
4
C 1
2
5
4
5
8
 3
1
4
C 1
2
5
D 4 3 C1 C

5
8

1
4
C

2
5

D 2 C
5 ð 5 10 ð 1 C8 ð 2
40
D 2 C
25  10 C16
40
D 2 C
31
40
D 2
31
40
Problem 4. Find the value of
3
7
ð
14
15
Dividing numerator and denominator by 3 gives:
1
3
7
ð
14
15
5
D

1
7
ð
14
5
D
1 ð 14
7 ð 5
Dividing numerator and denominator by 7 gives:
1 ð 14
2
1
7 ð 5
D
1 ð 2
1 ð 5
D
2
5
This process of dividing both the numerator and
denominator of a fraction by the same factor(s) is
called cancelling.
Problem 5. Evaluate 1
3
5
ð 2
1
3
ð 3
3

7
Mixed numbers must be expressed as improper
fractions before multiplication can be performed.
Thus,
1
3
5
ð 2
1
3
ð 3
3
7
D

5
5
C
3
5

ð

6
3
C
1
3

ð


21
7
C
3
7

D
8
5
ð
1
7
1
3
ð
24
8
7
1
D
8 ð 1 ð8
5 ð 1 ð1
D
64
5
D 12
4
5
Problem 6. Simplify

3
7
ł
12
21
3
7
ł
12
21
D
3
7
12
21
Multiplying both numerator and denominator by the
reciprocal of the denominator gives:
3
7
12
21
D
1
3
1
7
ð
21
3
12

4
1
12
1
21
ð
21
1
12
1
D
3
4
1
D
3
4
This method can be remembered by the rule: invert
the second fraction and change the operation from
division to multiplication. Thus:
3
7
ł
12
21
D
1
3
1
7

ð
21
3
12
4
D
3
4
as obtained previously.
Problem 7. Find the value of 5
3
5
ł 7
1
3
The mixed numbers must be expressed as improper
fractions. Thus,
5
3
5
ł 7
1
3
D
28
5
ł
22
3
D

14
28
5
ð
3
22
11
D
42
55
Problem 8. Simplify
1
3


2
5
C
1
4

ł

3
8
ð
1
3

The order of precedence of operations for problems

containing fractions is the same as that for inte-
gers, i.e. remembered by BODMAS (Brackets, Of,
Division, Multiplication, Addition and Subtraction).
Thus,
1
3


2
5
C
1
4

ł

3
8
ð
1
3

REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 3
D
1
3

4 ð 2 C5 ð 1
20
ł

3
1
24
8
(B)
D
1
3

13
5
20
ð
8
2
1
(D)
D
1
3

26
5
(M)
D
5 ð1  3 ð26
15
(S)
D
73

15
D
−4
13
15
Problem 9. Determine the value of
7
6
of

3
1
2
 2
1
4

C 5
1
8
ł
3
16

1
2
7
6
of


3
1
2
 2
1
4

C 5
1
8
ł
3
16

1
2
D
7
6
of 1
1
4
C
41
8
ł
3
16

1

2
(B)
D
7
6
ð
5
4
C
41
8
ł
3
16

1
2
(O)
D
7
6
ð
5
4
C
41
1
8
ð
16

2
3

1
2
(D)
D
35
24
C
82
3

1
2
(M)
D
35 C 656
24

1
2
(A)
D
691
24

1
2
(A)

D
691  12
24
(S)
D
679
24
D 28
7
24
Now try the following exercise
Exercise 1 Further problems on fractions
Evaluate the following:
1. (a)
1
2
C
2
5
(b)
7
16

1
4

(a)
9
10
(b)

3
16

2. (a)
2
7
C
3
11
(b)
2
9

1
7
C
2
3

(a)
43
77
(b)
47
63

3. (a) 10
3
7
 8

2
3
(b) 3
1
4
 4
4
5
C 1
5
6

(a) 1
16
21
(b)
17
60

4. (a)
3
4
ð
5
9
(b)
17
35
ð
15

119

(a)
5
12
(b)
3
49

5. (a)
3
5
ð
7
9
ð 1
2
7
(b)
13
17
ð 4
7
11
ð 3
4
39

(a)
3

5
(b) 11

6. (a)
3
8
ł
45
64
(b) 1
1
3
ł 2
5
9

(a)
8
15
(b)
12
23

7.
1
2
C
3
5
ł

8
15

1
3

1
7
24

8.
7
15
of

15 ð
5
7

C

3
4
ł
15
16

5
4
5


9.
1
4
ð
2
3

1
3
ł
3
5
C
2
7


13
126

10.

2
3
ð 1
1
4

ł


2
3
C
1
4

C 1
3
5

2
28
55

1.2 Ratio and proportion
The ratio of one quantity to another is a fraction, and
is the number of times one quantity is contained in
another quantity of the same kind. If one quantity is
directly proportional to another, then as one quan-
tity doubles, the other quantity also doubles. When a
quantity is inversely proportional to another, then
as one quantity doubles, the other quantity is halved.
Problem 10. A piece of timber 273 cm
long is cut into three pieces in the ratio of 3
to 7 to 11. Determine the lengths of the three
pieces
4 ENGINEERING MATHEMATICS
The total number of parts is 3 C7 C 11, that is, 21.
Hence 21 parts correspond to 273 cm

1 part corresponds to
273
21
D 13 cm
3 parts correspond to 3 ð13 D 39 cm
7 parts correspond to 7 ð13 D 91 cm
11 parts correspond to 11 ð13 D 143 cm
i.e. the lengths of the three pieces are 39 cm,
91 cm and 143 cm.
(Check: 39 C 91 C143 D 273)
Problem 11. A gear wheel having 80 teeth
is in mesh with a 25 tooth gear. What is the
gear ratio?
Gear ratio D 80:25 D
80
25
D
16
5
D 3.2
i.e. gear ratio D 16 : 5 or 3.2 : 1
Problem 12. An alloy is made up of
metals A and B in the ratio 2.5:1bymass.
How much of A has to be added to 6 kg of
B to make the alloy?
Ratio A : B: :2.5:1(i.e.AistoBas2.5isto1)
or
A
B
D

2.5
1
D 2.5
When B D 6 kg,
A
6
D 2.5 from which,
A D 6 ð2.5 D 15 kg
Problem 13. If 3 people can complete a
task in 4 hours, how long will it take 5
people to complete the same task, assuming
the rate of work remains constant
The more the number of people, the more quickly
the task is done, hence inverse proportion exists.
3 people complete the task in 4 hours,
1 person takes three times as long, i.e.
4 ð 3 D 12 hours,
5 people can do it in one fifth of the time that
one person takes, that is
12
5
hours or 2hours
24 minutes.
Now try the following exercise
Exercise 5 Further problems on ratio and
proportion
1. Divide 621 cm in the ratio of 3 to 7 to 13.
[81 cm to 189 cm to 351 cm]
2. When mixing a quantity of paints, dyes of
four different colours are used in the ratio

of 7:3:19:5. If the mass of the first dye
used is 3
1
2
g, determine the total mass of
the dyes used. [17 g]
3. Determine how much copper and how
much zinc is needed to make a 99 kg
brass ingot if they have to be in the
proportions copper : zinc: :8 : 3 by mass.
[72 kg : 27 kg]
4. It takes 21 hours for 12 men to resurface
a stretch of road. Find how many men
it takes to resurface a similar stretch of
road in 50 hours 24 minutes, assuming
the work rate remains constant. [5]
5. It takes 3 hours 15 minutes to fly from
city A to city B at a constant speed. Find
how long the journey takes if
(a) the speed is 1
1
2
times that of the
original speed and
(b) if the speed is three-quarters of the
original speed.
[(a)2h10min(b)4h20min]
1.3 Decimals
The decimal system of numbers is based on the
digits 0 to 9. A number such as 53.17 is called

a decimal fraction, a decimal point separating the
integer part, i.e. 53, from the fractional part, i.e. 0.17
REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 5
A number which can be expressed exactly as
a decimal fraction is called a terminating deci-
mal and those which cannot be expressed exactly
as a decimal fraction are called non-terminating
decimals. Thus,
3
2
D 1.5 is a terminating decimal,
but
4
3
D 1.33333 is a non-terminating decimal.
1.33333 can be written as 1.
P
3, called ‘one point-
three recurring’.
The answer to a non-terminating decimal may be
expressed in two ways, depending on the accuracy
required:
(i) correct to a number of significant figures,that
is, figures which signify something, and
(ii) correct to a number of decimal places,thatis,
the number of figures after the decimal point.
The last digit in the answer is unaltered if the next
digit on the right is in the group of numbers 0, 1,
2, 3 or 4, but is increased by 1 if the next digit
on the right is in the group of numbers 5, 6, 7, 8

or 9. Thus the non-terminating decimal 7.6183
becomes 7.62, correct to 3 significant figures, since
the next digit on the right is 8, which is in the group
of numbers 5, 6, 7, 8 or 9. Also 7.6183 becomes
7.618, correct to 3 decimal places, since the next
digit on the right is 3, which is in the group of
numbers 0, 1, 2, 3 or 4.
Problem 14. Evaluate
42.7 C 3.04 C8.7 C 0.06
The numbers are written so that the decimal points
are under each other. Each column is added, starting
from the right.
42.7
3.04
8.7
0.06
54.50
Thus 42.7 Y 3.04 Y 8.7 Y 0.06 = 54.50
Problem 15. Take 81.70 from 87.23
The numbers are written with the decimal points
under each other.
87.23
81.70
5.53
Thus 87.23 − 81.70 = 5.53
Problem 16. Find the value of
23.4  17.83 57.6 C32.68
The sum of the positive decimal fractions is
23.4 C 32.68 D 56.08
The sum of the negative decimal fractions is

17.83 C 57. 6 D 75.43
Taking the sum of the negative decimal fractions
from the sum of the positive decimal fractions gives:
56.08  75. 43
i.e. 75.43 56.08 D
−19.35
Problem 17. Determine the value of
74.3 ð 3.8
When multiplying decimal fractions: (i) the numbers
are multiplied as if they are integers, and (ii) the
position of the decimal point in the answer is such
that there are as many digits to the right of it as the
sum of the digits to the right of the decimal points
of the two numbers being multiplied together. Thus
(i)
743
38
5 944
22 290
28 234
(ii) As there are 1 C1 D 2 digits to the right of
the decimal points of the two numbers being
multiplied together, (74.3
ð 3.8), then
74
.3 × 3.8 = 282.34
Problem 18. Evaluate 37.81 ł 1.7, correct
to (i) 4 significant figures and (ii) 4 decimal
places
6 ENGINEERING MATHEMATICS

37.81 ł 1.7 D
37.81
1.7
The denominator is changed into an integer by
multiplying by 10. The numerator is also multiplied
by 10 to keep the fraction the same. Thus
37.81 ł 1.7 D
37.81 ð 10
1.7 ð 10
D
378.1
17
The long division is similar to the long division of
integers and the first four steps are as shown:
22.24117
17

378.100000
34
38
34
41
34
70
68
20
(i) 37
.81 ÷ 1.7 = 22.24, correct to 4 significant
figures,and
(ii) 37

.81 ÷ 1.7 = 22.2412, correct to 4 decimal
places.
Problem 19. Convert (a) 0.4375 to a proper
fraction and (b) 4.285 to a mixed number
(a) 0.4375 can be written as
0.4375 ð10 000
10 000
without changing its value,
i.e. 0.4375 D
4375
10 000
By cancelling
4375
10 000
D
875
2000
D
175
400
D
35
80
D
7
16
i.e. 0
.4375 =
7
16

(b) Similarly, 4
.285 D 4
285
1000
D 4
57
200
Problem 20. Express as decimal fractions:
(a)
9
16
and (b) 5
7
8
(a) To convert a proper fraction to a decimal frac-
tion, the numerator is divided by the denomi-
nator. Division by 16 can be done by the long
division method, or, more simply, by dividing
by 2 and then 8:
4.50
2

9.00
0.5625
8

4.5000
Thus,
9
16

= 0.5625
(b) For mixed numbers, it is only necessary to
convert the proper fraction part of the mixed
number to a decimal fraction. Thus, dealing
with the
7
8
gives:
0.875
8

7.000
i.e.
7
8
D 0.875
Thus 5
7
8
= 5.875
Now try the following exercise
Exercise 3 Further problems on decimals
In Problems 1 to 6, determine the values of
the expressions given:
1. 23.6 C 14.71 18.9  7.421 [11.989]
2. 73.84  113.247 C8.21  0.068
[31.265]
3. 3.8 ð 4.1 ð0.7 [10.906]
4. 374.1 ð 0.006 [2.2446]
5. 421.8 ł 17, (a) correct to 4 significant

figures and (b) correct to 3 decimal
places.
[(a) 24.81 (b) 24.812]
6.
0.0147
2.3
, (a) correct to 5 decimal places
and (b) correct to 2 significant figures.
[(a) 0.00639 (b) 0.0064]
REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 7
7. Convert to proper fractions:
(a) 0.65 (b) 0.84 (c) 0.0125 (d) 0.282
and (e) 0.024

(a)
13
20
(b)
21
25
(c)
1
80
(d)
141
500
(e)
3
125


8. Convert to mixed numbers:
(a) 1.82 (b) 4.275 (c) 14.125 (d) 15.35
and (e) 16.2125




(a) 1
41
50
(b) 4
11
40
(c) 14
1
8
(d) 15
7
20
(e) 16
17
80




In Problems 9 to 12, express as decimal frac-
tions to the accuracy stated:
9.
4

9
, correct to 5 significant figures.
[0.44444]
10.
17
27
, correct to 5 decimal place.
[0.62963]
11. 1
9
16
, correct to 4 significant figures.
[1.563]
12. 13
31
37
, correct to 2 decimal places.
[13.84]
1.4 Percentages
Percentages areusedtogiveacommonstandard
and are fractions having the number 100 as their
denominators. For example, 25 per cent means
25
100
i.e.
1
4
and is written 25%.
Problem 21. Express as percentages:
(a) 1.875 and (b) 0.0125

A decimal fraction is converted to a percentage by
multiplying by 100. Thus,
(a) 1.875 corresponds to 1.875 ð 100%, i.e.
187.5%
(b) 0.0125 corresponds to 0.0125 ð 100%, i.e.
1.25%
Problem 22. Express as percentages:
(a)
5
16
and (b) 1
2
5
To convert fractions to percentages, they are (i) con-
verted to decimal fractions and (ii) multiplied by 100
(a) By division,
5
16
D 0.3125, hence
5
16
corre-
sponds to 0.3125 ð100%, i.e. 31.25%
(b) Similarly, 1
2
5
D 1.4 when expressed as a
decimal fraction.
Hence 1
2

5
D 1.4 ð100% D 140%
Problem 23. It takes 50 minutes to machine
a certain part. Using a new type of tool, the
time can be reduced by 15%. Calculate the
new time taken
15% of 50 minutes D
15
100
ð 50 D
750
100
D 7.5 minutes.
hence the new time taken is
50  7.5 D 42
.5 minutes.
Alternatively, if the time is reduced by 15%, then
it now takes 85% of the original time, i.e. 85% of
50 D
85
100
ð50 D
4250
100
D 42
.5 minutes, as above.
Problem 24. Find 12.5% of £378
12.5% of £378 means
12.5
100

ð 378, since per cent
means ‘per hundred’.
Hence 12.5% of £378 D
12.5
1
100
8
ð 378 D
1
8
ð 378 D
378
8
D £47
.25
8 ENGINEERING MATHEMATICS
Problem 25. Express 25 minutes as a
percentage of 2 hours, correct to the
nearest 1%
Working in minute units, 2 hours D 120 minutes.
Hence 25 minutes is
25
120
ths of 2 hours. By can-
celling,
25
120
D
5
24

Expressing
5
24
as a decimal fraction gives 0.208
P
3
Multiplying by 100 to convert the decimal fraction
to a percentage gives:
0.208
P
3 ð 100 D 20.8
P
3%
Thus 25 minutes is 21% of 2 hours, correct to the
nearest 1%.
Problem 26. A German silver alloy consists
of 60% copper, 25% zinc and 15% nickel.
Determine the masses of the copper, zinc and
nickel in a 3.74 kilogram block of the alloy
By direct proportion:
100% corresponds to 3.74 kg
1% corresponds to
3.74
100
D 0.0374 kg
60% corresponds to 60 ð 0.0374 D 2.244 kg
25% corresponds to 25 ð 0.0374 D 0.935 kg
15% corresponds to 15 ð 0.0374 D 0.561 kg
Thus, the masses of the copper, zinc and nickel are
2.244 kg, 0.935 kg and 0.561 kg, respectively.

(Check: 2.244 C0.935 C 0.561 D 3.74)
Now try the following exercise
Exercise 4 Further problems percentages
1. Convert to percentages:
(a) 0.057 (b) 0.374 (c) 1.285
[(a) 5.7% (b) 37.4% (c) 128.5%]
2. Express as percentages, correct to 3
significant figures:
(a)
7
33
(b)
19
24
(c) 1
11
16
[(a) 21.2% (b) 79.2% (c) 169%]
3. Calculate correct to 4 significant figures:
(a) 18% of 2758 tonnes (b) 47% of
18.42 grams (c) 147% of 14.1 seconds
[(a) 496.4 t (b) 8.657 g (c) 20.73 s]
4. When 1600 bolts are manufactured, 36
are unsatisfactory. Determine the percent-
age unsatisfactory. [2.25%]
5. Express: (a) 140 kg as a percentage of
1t (b)47s as a percentage of 5min
(c) 13.4 cm as a percentage of 2.5 m
[(a) 14% (b) 15.67% (c) 5.36%]
6. A block of monel alloy consists of 70%

nickel and 30% copper. If it contains
88.2 g of nickel, determine the mass of
copper in the block. [37.8 g]
7. A drilling machine should be set to
250 rev/min. The nearest speed available
on the machine is 268 rev/min. Calculate
the percentage over speed. [7.2%]
8. Two kilograms of a compound contains
30% of element A, 45% of element B and
25% of element C. Determine the masses
of the three elements present.
[A 0.6 kg, B 0.9 kg, C 0.5 kg]
9. A concrete mixture contains seven parts
by volume of ballast, four parts by vol-
ume of sand and two parts by volume of
cement. Determine the percentage of each
of these three constituents correct to the
nearest 1% and the mass of cement in a
two tonne dry mix, correct to 1 significant
figure.
[54%, 31%, 15%, 0.3 t]
2
Indices and standard form
2.1 Indices
The lowest factors of 2000 are 2ð2ð2ð2ð 5ð5ð5.
These factors are written as 2
4
ð5
3
, where 2 and 5

are called bases and the numbers 4 and 3 are called
indices.
When an index is an integer it is called a power.
Thus, 2
4
is called ‘two to the power of four’, and
has a base of 2 and an index of 4. Similarly, 5
3
is
called ‘five to the power of 3’ and has a base of 5
andanindexof3.
Special names may be used when the indices are
2 and 3, these being called ‘squared’ and ‘cubed’,
respectively. Thus 7
2
is called ‘seven squared’ and
9
3
is called ‘nine cubed’. When no index is shown,
the power is 1, i.e. 2 means 2
1
.
Reciprocal
The reciprocal of a number is when the index is
1 and its value is given by 1, divided by the base.
Thus the reciprocal of 2 is 2
1
and its value is
1
2

or 0.5. Similarly, the reciprocal of 5 is 5
1
which
means
1
5
or 0.2
Square root
The square root of a number is when the index is
1
2
,
and the square root of 2 is written as 2
1/2
or
p
2. The
value of a square root is the value of the base which
when multiplied by itself gives the number. Since
3ð3 D 9, then
p
9 D 3. However, 3ð3 D 9,
so
p
9 D3. There are always two answers when
finding the square root of a number and this is shown
by putting both a C and a  sign in front of the
answer to a square root problem. Thus
p
9 Dš3

and 4
1/2
D
p
4 Dš2, and so on.
Laws of indices
When simplifying calculations involving indices,
certain basic rules or laws can be applied, called
the laws of indices. These are given below.
(i) When multiplying two or more numbers hav-
ing the same base, the indices are added. Thus
3
2
ð 3
4
D 3
2C4
D 3
6
(ii) When a number is divided by a number having
the same base, the indices are subtracted. Thus
3
5
3
2
D 3
52
D 3
3
(iii) When a number which is raised to a power

is raised to a further power, the indices are
multiplied. Thus
3
5

2
D 3
5ð2
D 3
10
(iv) When a number has an index of 0, its value
is 1. Thus 3
0
D 1
(v) A number raised to a negative power is the
reciprocal of that number raised to a positive
power. Thus 3
4
D
1
3
4
Similarly,
1
2
3
D 2
3
(vi) When a number is raised to a fractional power
the denominator of the fraction is the root of

the number and the numerator is the power.
Thus 8
2/3
D
3
p
8
2
D 2
2
D 4
and 25
1/2
D
2
p
25
1
D
p
25
1
Dš5
(Note that
p
Á
2
p
)
2.2 Worked problems on indices

Problem 1. Evaluate: (a) 5
2
ð 5
3
,
(b) 3
2
ð 3
4
ð 3and(c)2ð2
2
ð 2
5
From law (i):
(a) 5
2
ð5
3
D 5
2C3
D 5
5
D 5ð5ð5ð5ð5 D 3125
10 ENGINEERING MATHEMATICS
(b) 3
2
ð 3
4
ð 3 D 3
2C4C1

D 3
7
D 3 ð3 ðÐÐÐ to7terms
D 2187
(c) 2 ð2
2
ð 2
5
D 2
1C2C5
D 2
8
D 256
Problem 2. Find the value of:
(a)
7
5
7
3
and (b)
5
7
5
4
From law (ii):
(a)
7
5
7
3

D 7
53
D 7
2
D 49
(b)
5
7
5
4
D 5
74
D 5
3
D 125
Problem 3. Evaluate: (a) 5
2
ð 5
3
ł 5
4
and
(b) 3 ð 3
5
 ł 3
2
ð 3
3

From laws (i) and (ii):

(a) 5
2
ð 5
3
ł 5
4
D
5
2
ð 5
3
5
4
D
5
2C3
5
4
D
5
5
5
4
D 5
54
D 5
1
D 5
(b) 3 ð 3
5

 ł 3
2
ð 3
3
 D
3 ð 3
5
3
2
ð 3
3
D
3
1C5
3
2C3
D
3
6
3
5
D 3
65
D 3
1
D 3
Problem 4. Simplify: (a) 2
3

4

(b) 3
2

5
,
expressing the answers in index form.
From law (iii):
(a) 2
3

4
D 2
3ð4
D 2
12
(b) 3
2

5
D 3
2ð5
D 3
10
Problem 5. Evaluate:
10
2

3
10
4

ð 10
2
From the laws of indices:
10
2

3
10
4
ð 10
2
D
10
2ð3
10
4C2
D
10
6
10
6
D 10
66
D 10
0
D 1
Problem 6. Find the value of
(a)
2
3

ð 2
4
2
7
ð 2
5
and (b)
3
2

3
3 ð 3
9
From the laws of indices:
(a)
2
3
ð 2
4
2
7
ð 2
5
D
2
3C4
2
7C5
D
2

7
2
12
D 2
712
D 2
5
D
1
2
5
D
1
32
(b)
3
2

3
3 ð 3
9
D
3
2ð3
3
1C9
D
3
6
3

10
D 3
610
D 3
4
D
1
3
4
D
1
81
Now try the following exercise
Exercise 5 Further problems on indices
In Problems 1 to 10, simplify the expressions
given, expressing the answers in index form
and with positive indices:
1. (a) 3
3
ð 3
4
(b) 4
2
ð 4
3
ð 4
4
[(a) 3
7
(b) 4

9
]
2. (a) 2
3
ð 2 ð2
2
(b) 7
2
ð 7
4
ð 7 ð7
3
[(a) 2
6
(b) 7
10
]
3. (a)
2
4
2
3
(b)
3
7
3
2
[(a) 2 (b) 3
5
]

4. (a) 5
6
ł 5
3
(b) 7
13
/7
10
[(a) 5
3
(b) 7
3
]
5. (a) 7
2

3
(b) 3
3

2
[(a) 7
6
(b) 3
6
]
6. (a)
2
2
ð 2

3
2
4
(b)
3
7
ð 3
4
3
5
[(a) 2 (b) 3
6
]
7. (a)
5
7
5
2
ð 5
3
(b)
13
5
13 ð 13
2
[(a) 5
2
(b) 13
2
]

8. (a)
9 ð3
2

3
3 ð27
2
(b)
16 ð4
2
2 ð8
3
[(a) 3
4
(b) 1]
INDICES AND STANDARD FORM 11
9. (a)
5
2
5
4
(b)
3
2
ð 3
4
3
3

(a) 5

2
(b)
1
3
5

10. (a)
7
2
ð 7
3
7 ð 7
4
(b)
2
3
ð 2
4
ð 2
5
2 ð 2
2
ð 2
6

(a) 7
2
(b)
1
2


2.3 Further worked problems on
indices
Problem 7. Evaluate:
3
3
ð 5
7
5
3
ð 3
4
The laws of indices only apply to terms having the
same base. Grouping terms having the same base,
and then applying the laws of indices to each of the
groups independently gives:
3
3
ð 5
7
5
3
ð 3
4
D
3
3
3
4
ð

5
7
5
3
D 3
34
ð 5
73
D 3
1
ð 5
4
D
5
4
3
1
D
625
3
D 208
1
3
Problem 8. Find the value of
2
3
ð 3
5
ð 7
2


2
7
4
ð 2
4
ð 3
3
2
3
ð 3
5
ð 7
2

2
7
4
ð 2
4
ð 3
3
D 2
34
ð 3
53
ð 7
2ð24
D 2
1

ð 3
2
ð 7
0
D
1
2
ð 3
2
ð 1 D
9
2
D 4
1
2
Problem 9. Evaluate:
(a) 4
1/2
(b) 16
3/4
(c) 27
2/3
(d) 9
1/2
(a) 4
1/2
D
p
4 D ±2
(b) 16

3/4
D
4
p
16
3
D š2
3
D ±8
(Note that it does not matter whether the 4th root
of 16 is found first or whether 16 cubed is found
first — the same answer will result).
(c) 27
2/3
D
3
p
27
2
D 3
2
D 9
(d) 9
1/2
D
1
9
1/2
D
1

p
9
D
1
š3
D
±
1
3
Problem 10. Evaluate:
4
1.5
ð 8
1/3
2
2
ð 32
2/5
4
1.5
D 4
3/2
D
p
4
3
D 2
3
D 8,
8

1/3
D
3
p
8 D 2, 2
2
D 4
and 32
2/5
D
1
32
2/5
D
1
5
p
32
2
D
1
2
2
D
1
4
Hence
4
1.5
ð 8

1/3
2
2
ð 32
2/5
D
8 ð 2
4 ð
1
4
D
16
1
D 16
Alternatively,
4
1.5
ð 8
1/3
2
2
ð 32
2/5
D
[2
2
]
3/2
ð 2
3


1/3
2
2
ð 2
5

2/5
D
2
3
ð 2
1
2
2
ð 2
2
D 2
3C122
D 2
4
D 16
Problem 11. Evaluate:
3
2
ð 5
5
C 3
3
ð 5

3
3
4
ð 5
4
Dividing each term by the HCF (i.e. highest com-
mon factor) of the three terms, i.e. 3
2
ð 5
3
,gives:
3
2
ð 5
5
C 3
3
ð 5
3
3
4
ð 5
4
D
3
2
ð 5
5
3
2

ð 5
3
C
3
3
ð 5
3
3
2
ð 5
3
3
4
ð 5
4
3
2
ð 5
3
D
3
22
ð 5
53
C 3
32
ð 5
0
3
42

ð 5
43
D
3
0
ð 5
2
C 3
1
ð 5
0
3
2
ð 5
1
D
1 ð 25 C3 ð 1
9 ð 5
D
28
45
Problem 12. Find the value of
3
2
ð 5
5
3
4
ð 5
4

C 3
3
ð 5
3
12 ENGINEERING MATHEMATICS
To simplify the arithmetic, each term is divided by
the HCF of all the terms, i.e. 3
2
ð 5
3
. Thus
3
2
ð 5
5
3
4
ð 5
4
C 3
3
ð 5
3
D
3
2
ð 5
5
3
2

ð 5
3
3
4
ð 5
4
3
2
ð 5
3
C
3
3
ð 5
3
3
2
ð 5
3
D
3
22
ð 5
53
3
42
ð 5
43
C 3
32

ð 5
33
D
3
0
ð 5
2
3
2
ð 5
1
C 3
1
ð 5
0
D
25
45 C 3
D
25
48
Problem 13. Simplify:

4
3

3
ð

3

5

2

2
5

3
giving the answer with positive indices
A fraction raised to a power means that both the
numerator and the denominator of the fraction are
raised to that power, i.e.

4
3

3
D
4
3
3
3
A fraction raised to a negative power has the
same value as the inverse of the fraction raised to a
positive power.
Thus,

3
5


2
D
1

3
5

2
D
1
3
2
5
2
D 1 ð
5
2
3
2
D
5
2
3
2
Similarly,

2
5

3

D

5
2

3
D
5
3
2
3
Thus,

4
3

3
ð

3
5

2

2
5

3
D
4

3
3
3
ð
5
2
3
2
5
3
2
3
D
4
3
3
3
ð
5
2
3
2
ð
2
3
5
3
D
2
2


3
ð 2
3
3
3C2
ð 5
32
D
2
9
3
5
× 5
Now try the following exercise
Exercise 6 Further problems on indices
In Problems 1 and 2, simplify the expressions
given, expressing the answers in index form
and with positive indices:
1. (a)
3
3
ð 5
2
5
4
ð 3
4
(b)
7

2
ð 3
2
3
5
ð 7
4
ð 7
3

(a)
1
3 ð 5
2
(b)
1
7
3
ð 3
7

2. (a)
4
2
ð 9
3
8
3
ð 3
4

(b)
8
2
ð 5
2
ð 3
4
25
2
ð 2
4
ð 9
2

(a)
3
2
2
5
(b)
1
2
10
ð 5
2

3. Evaluate a

1
3

2

1
b 81
0.25
c 16
1/4
d

4
9

1/2

(a) 9 (b) š3(c)š
1
2
(d) š
2
3

In Problems 4 to 8, evaluate the expressions
given.
4.
9
2
ð 7
4
3
4

ð 7
4
C 3
3
ð 7
2

147
148

5.
2
4

2
 3
2
ð 4
4
2
3
ð 16
2

1
9

6.

1

2

3


2
3

2

3
5

2

5
65
72

7.

4
3

4

2
9

2

[64]
8.
3
2

3/2
ð 8
1/3

2
3
2
ð 4
3

1/2
ð 9
1/2

4
1
2

INDICES AND STANDARD FORM 13
2.4 Standard form
A number written with one digit to the left of the
decimal point and multiplied by 10 raised to some
power is said to be written in standard form. Thus:
5837 is written as 5.837 ð 10
3

in standard form,
and 0.0415 is written as 4.15 ð 10
2
in standard
form.
When a number is written in standard form, the
first factor is called the mantissa and the second
factor is called the exponent. Thus the number
5.8 ð 10
3
has a mantissa of 5.8 and an exponent
of 10
3
.
(i) Numbers having the same exponent can be
added or subtracted in standard form by adding
or subtracting the mantissae and keeping the
exponent the same. Thus:
2.3 ð 10
4
C 3.7 ð10
4
D 2.3 C3.7 ð 10
4
D 6.0 ð10
4
and 5.9 ð10
2
 4.6 ð10
2

D 5.9 4.6 ð 10
2
D 1.3 ð10
2
When the numbers have different exponents,
one way of adding or subtracting the numbers
is to express one of the numbers in non-
standard form, so that both numbers have the
same exponent. Thus:
2.3 ð 10
4
C 3.7 ð10
3
D 2.3 ð10
4
C 0.37 ð10
4
D 2.3 C0.37 ð 10
4
D 2.67 ð10
4
Alternatively,
2.3 ð 10
4
C 3.7 ð10
3
D 23 000 C3700 D 26 700
D 2.67 ð10
4
(ii) The laws of indices are used when multiplying

or dividing numbers given in standard form.
For example,
2.5 ð10
3
 ð 5 ð10
2

D 2.5 ð 5 ð 10
3C2

D 12.5 ð10
5
or 1.25 ð10
6
Similarly,
6 ð 10
4
1.5 ð 10
2
D
6
1.5
ð 10
42
 D 4 ð 10
2
2.5 Worked problems on standard
form
Problem 14. Express in standard form:
(a) 38.71 (b) 3746 (c) 0.0124

For a number to be in standard form, it is expressed
with only one digit to the left of the decimal point.
Thus:
(a) 38.71 must be divided by 10 to achieve one
digit to the left of the decimal point and it
must also be multiplied by 10 to maintain the
equality, i.e.
38.71 D
38.71
10
ð10 D 3
.871 × 10 in standard
form
(b) 3746 D
3746
1000
ð 1000 D 3
.746 × 10
3
in stan-
dard form
(c) 0.0124 D 0.0124 ð
100
100
D
1.24
100
D 1
.24 × 10
−2

in standard form
Problem 15. Express the following
numbers, which are in standard form, as
decimal numbers: (a) 1.725 ð10
2
(b) 5.491 ð10
4
(c) 9.84 ð10
0
(a) 1.725 ð10
2
D
1.725
100
D 0
.01725
(b) 5.491 ð10
4
D 5.491 ð10 000 D 54 910
(c) 9.84 ð 10
0
D 9.84 ð1 D 9.84 (since 10
0
D 1)
Problem 16. Express in standard form,
correct to 3 significant figures:
(a)
3
8
(b) 19

2
3
(c) 741
9
16